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I know this has been asked a lot, but I haven't found a good solution:

Show that the set $\{1, x, x^2, ..., x^n\}$ constitutes a basis of the vector space of polynomial functions $\varphi : \mathbb R \to \mathbb R$.

My problem is to proof linear independency.
Suppose $$\sum_{i=0}^na_ix^i = 0$$ for $a_i \in \mathbb R. $ From plugging in $x = 0$ I get $a_0 = 0$. But how to proceed from here? I know I can factorize $x$ like this: $$x(a_1 + a_2x + ... + a_nx^{n-1}) = 0$$ and for $x\neq 0$ it must be $$a_1 + a_2x + ... + a_nx^{n-1} = 0$$ but here I am stuck. I would like to proof this with basic algebra and possibly without theorems from which this easily follows.

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    The question is imprecise: did you intend the set to be $\{1,x,x^2,\ldots,x^n,\ldots\}$? Or did you mean the vector space of polynomials of degree $\le n$? – 2017-01-12
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    The vector space is defined as "Vector space of polynomial functions $V := \{\varphi: \mathbb R \to \mathbb R\, |\, \exists n \in \mathbb N, a_0, a_1, ... , a_n \in \mathbb R: \varphi (t) = \sum_{i=0}^n a_i t^i \text{ for all } t \in \mathbb R\}$". So I suppose it should've been $\{1,x,x^2,...\}$ – 2017-01-12

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"From plugging in $x= 0$, I get $a_0= 0$."

Good, now differentiate $\sum_{i=0}^n a_ix^i$ to get $\sum_{i=1}^n ia_ix^{i-1}$. Setting $j=i-1$, $i=j+1$, when $i=1$, $j=0$ and when $i=n$, $j=n- 1$ so this becomes $\sum_{j=0}^{n-1} (j+1)a_{j+1}x^j$. Now plugging in $x=0$, $a_1=0$. Do that $n$ times to show that all $n$ coefficients are $0$.

(That is why what Jorge Fernandez Hidalgo says is correct.)

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    I don't understand why you need to do that, the coefficients of the zero polynomial are all $0$. Polynomials are "formal" objects, you don't need to use properties of their interpretation as functions $\mathbb R \rightarrow \mathbb R$. – 2017-01-11
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    @JorgeFernándezHidalgo: This question specifically asks about polynomials as functions $\mathbb{R}\to\mathbb{R}$, not as formal expressions. – 2017-01-11
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    Eric is right, I was talking about the functions $f:\mathbb R \to \mathbb R$. The correct solution was this one. I should've been more clear though, so it's fine, now I learned about solutions for both the polynomials as formal objects and as real-valued functions. :) – 2017-01-13
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Suppose $f(x)=\sum_{k=0}^n a_kx^k=0$ for all $x\in\Bbb R$ and let $m$ be the least integer such that $a_m\ne 0$. Then, $$0=\lim_{x\to 0} \frac{f(x)}{x^m}=\lim_{x\to 0}\ a_m+\sum_{k=1}^{n-m} a_{k+m}x^k=a_m$$

which is absurd.